CodeForces 554B Ohana Cleans Up

Ohana Cleans Up
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice CodeForces
554B

Description

Ohana Matsumae is trying to clean a room, which is divided up into an n by n grid of squares. Each square is initially either
clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room
to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.

Return the maximum number of rows that she can make completely clean.

Input

The first line of input will be a single integer n (1 ≤ n ≤ 100).

The next n lines will describe the state of the room. The i-th line will contain a binary string with n characters
denoting the state of the i-th row of the room. The j-th character on this line is '1' if the j-th
square in the i-th row is clean, and '0' if it is dirty.

Output

The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.

Sample Input

Input

4
0101
1000
1111
0101

Output

2

Input

3
111
111
111

Output

3

Hint

In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.

In the second sample, everything is already clean, so Ohana doesn't need to do anything.

题意：一个房间 n*n ,有脏的地方也有干净的地方，脏的地方记为0，干净的地方记为1， 已知 干净的地方擦过一遍就脏了，脏的地方擦过一遍就干净了。 要求每次要擦就擦一列，要不就不擦，最后输出干净的行数，同时行数是最大的。

输入：一个数 n 代表有几行几列，随后输入 n * n 的矩阵。

题解： 因为每次都要擦一列， 所以一列中，开始时相同的状态始终状态一样， 开始时不同的状态始终不一样，所以要求干净的行数最大，我们就求出相同的有多少行就行了，当然要记录其中的最大值。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
char s[150][150];
int vis[150];

int main(){
int n, i;
while(~scanf("%d", &n)){
for(i = 0; i < n; ++i){
scanf("%s", &s[i]);
}
memset(vis, 0, sizeof(vis));

/*for(i = 0; i < n; ++i)
cout << s[i] << endl;*/
int result = 1;
int maxn = 0;
for(i = 0; i < n; ++i){
result = 1;
for(int j = i + 1; j < n; ++j){
if(vis[i] == 0 && strcmp(s[i], s[j]) == 0){
++result;
vis[j] = 1;
}
}
maxn = max(result, maxn);
}
cout << maxn << endl;
}
return 0;
}